Problem: Consider the polar curve $r=4\sin(\theta)$. What is the equation of the tangent line to the curve $r$ at $\theta=\dfrac{\pi}{6}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y+1=\sqrt{3}\left(x+\sqrt{3}\right)$ (Choice B) B $y-1=\sqrt{3}\left(x-\sqrt{3}\right)$ (Choice C) C $y+1=\dfrac{1}{\sqrt{3}}\left(x+\sqrt{3}\right)$ (Choice D) D $y-1=\dfrac{1}{\sqrt{3}}\left(x-\sqrt{3}\right)$
Explanation: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Then we can use the point-slope form to complete the equation for the tangent line through $\theta=\dfrac{\pi}{6}$. For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={4\sin(\theta)}\cos(\theta) \\\\ &=2\sin(2\theta) \\\\ y&={4\sin(\theta)} \sin(\theta) \\\\ &=4\sin^2(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{4\sin\left(2\theta\right) }{4\cos\left(2\theta\right) } \\\\ &=\tan\left(2\theta\right) \end{aligned}$ Evaluating $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{6}}$ gives us the slope of our tangent line. $\begin{aligned} {\left. \dfrac{dy}{dx}\right| _{\theta=\tfrac{\pi}{6}}}&=\tan\left(2\left({\dfrac{\pi}{6}}\right)\right) \\\\ &=\tan\left(\dfrac{\pi}{3}\right) \\\\ &={\sqrt{3}} \end{aligned}$ We now find $x$ and $y$ at the point $\theta=\dfrac{\pi}{6}$. $\begin{aligned} {x\left({\dfrac{\pi}{6}}\right)}&=2\sin\left(2\left({\dfrac{\pi}{6}}\right)\right) \\\\ &=2\sin\left(\dfrac{\pi}{3}\right) \\\\ &=2\left(\dfrac{\sqrt{3}}{2}\right) \\\\ &={\sqrt{3}} \\\\ \\\\ y\left({\dfrac{\pi}{6}}\right)}&=4\sin^2\left({\dfrac{\pi}{6}}\right) \\\\ &=4\left(\dfrac{1}{2}\right)^2 \\\\ &=1} \end{aligned}$ Therefore the equation of our tangent line is: $y-1}={\sqrt{3}}\left(x-{\sqrt{3}}\right)$ The graph of the tangent is shown.